【题目】

Given a binary tree, return the preorder traversal of its nodes‘ values.

For example:
Given binary tree {1,#,2,3},

   1
         2
    /
   3

return [1,2,3].

Note: Recursive solution is trivial, could you do it iteratively?

【代码一】

/*********************************
*   日期：2014-10-15
*   作者：SJF0115
*   题号: Binary Tree Preorder Traversal
*   来源：https://oj.leetcode.com/problems/binary-tree-preorder-traversal/
*   结果：AC
*   来源：LeetCode
*   总结：
**********************************/
#include <iostream>
#include <malloc.h>
#include <vector>
using namespace std;

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

class Solution {
public:
    vector<int> v;
    void PreOrder(TreeNode *root){
        if (root == NULL){
            return;
        }
        v.push_back(root->val);
        PreOrder(root->left);
        PreOrder(root->right);
    }
    vector<int> preorderTraversal(TreeNode *root) {
        PreOrder(root);
        return v;
    }
};

//按先序序列创建二叉树
int CreateBTree(TreeNode* &T){
    char data;
    //按先序次序输入二叉树中结点的值（一个字符），‘#’表示空树
    cin>>data;
    if(data =http://www.mamicode.com/= '#'){>
【代码二】
非递归实现
class Solution {
public:
    vector<int> preorderTraversal(TreeNode *root) {
        vector<int> v;
        stack<TreeNode*> s;
        TreeNode* p = root;
        //栈不空或者p不空时循环
        while(p != NULL || !s.empty()){
            if(p != NULL){
                //访问根节点
                v.push_back(p->val);
                //根节点插入栈中,用来访问右子树
                s.push(p);
                //遍历左子树
                p = p->left;
            }
            else{
                //左子树访问完毕，访问右子树
                p = s.top();
                s.pop();
                p = p->right;
            }
        }
        return v;
    }
};








[LeetCode]Binary Tree Preorder Traversal